Is csc^2(x)-1=(cos(x))/(tan(x)sin(x)) ?

The answer to whether csc^2(x)-1=(cos(x))/(tan(x)sin(x)) is True

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Popular Trigonometry >

prove csc^2(x)-1=(cos(x))/(tan(x)sin(x))

Solution

prove

csc^{2} (x) - 1 = \frac{cos ( x )}{tan ( x ) sin ( x )}

Solution

True

Solution steps

csc^{2} (x) - 1 = \frac{cos ( x )}{tan ( x ) sin ( x )}

Manipulating right side

\frac{cos ( x )}{tan ( x ) sin ( x )}

Express with sin, cos

\frac{cos ( x )}{sin ( x ) tan ( x )}

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{cos ( x )}{sin ( x ) \frac{s i n ( x )}{c o s ( x )}}

Simplify

\frac{cos ( x )}{sin ( x ) \frac{s i n ( x )}{c o s ( x )}} : \frac{cos ^{2} ( x )}{sin ^{2} ( x )}

\frac{cos ( x )}{sin ( x ) \frac{s i n ( x )}{c o s ( x )}}

Multiply

sin (x) \frac{sin ( x )}{cos ( x )} : \frac{sin ^{2} ( x )}{cos ( x )}

sin (x) \frac{sin ( x )}{cos ( x )}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{sin ( x ) sin ( x )}{cos ( x )}

sin (x) sin (x) = sin^{2} (x)

sin (x) sin (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin (x) sin (x) = sin^{1 + 1} (x)

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

= \frac{sin ^{2} ( x )}{cos ( x )}

= \frac{cos ( x )}{\frac{s i n ^{2} ( x )}{c o s ( x )}}

Apply the fraction rule:

\frac{a}{\frac{b}{c}} = \frac{a \cdot c}{b}

= \frac{cos ( x ) cos ( x )}{sin ^{2} ( x )}

cos (x) cos (x) = cos^{2} (x)

cos (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (x) cos (x) = cos^{1 + 1} (x)

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

= \frac{cos ^{2} ( x )}{sin ^{2} ( x )}

= \frac{cos ^{2} ( x )}{sin ^{2} ( x )}

= \frac{cos ^{2} ( x )}{sin ^{2} ( x )}

Rewrite using trig identities

\frac{cos ^{2} ( x )}{sin ^{2} ( x )}

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= \frac{1 - sin ^{2} ( x )}{sin ^{2} ( x )}

= \frac{1 - sin ^{2} ( x )}{sin ^{2} ( x )}

Rewrite using trig identities

Use the basic trigonometric identity:

sin (x) = \frac{1}{csc ( x )}

\frac{1 - ( \frac{1}{c s c ( x )} ) ^{2}}{( \frac{1}{c s c ( x )} ) ^{2}}

Simplify

\frac{1 - ( \frac{1}{c s c ( x )} ) ^{2}}{( \frac{1}{c s c ( x )} ) ^{2}}

(\frac{1}{csc ( x )})^{2} = \frac{1}{csc ^{2} ( x )}

(\frac{1}{csc ( x )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{csc ^{2} ( x )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{csc ^{2} ( x )}

= \frac{1 - ( \frac{1}{c s c ( x )} ) ^{2}}{\frac{1}{c s c ^{2} ( x )}}

(\frac{1}{csc ( x )})^{2} = \frac{1}{csc ^{2} ( x )}

(\frac{1}{csc ( x )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{csc ^{2} ( x )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{csc ^{2} ( x )}

= \frac{1 - \frac{1}{c s c ^{2} ( x )}}{\frac{1}{c s c ^{2} ( x )}}

Apply the fraction rule:

\frac{a}{\frac{b}{c}} = \frac{a \cdot c}{b}

= \frac{( 1 - \frac{1}{c s c ^{2} ( x )} ) csc ^{2} ( x )}{1}

Join

1 - \frac{1}{csc ^{2} ( x )} : \frac{csc ^{2} ( x ) - 1}{csc ^{2} ( x )}

1 - \frac{1}{csc ^{2} ( x )}

Convert element to fraction:

1 = \frac{1 csc ^{2} ( x )}{csc ^{2} ( x )}

= \frac{1 \cdot csc ^{2} ( x )}{csc ^{2} ( x )} - \frac{1}{csc ^{2} ( x )}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 \cdot csc ^{2} ( x ) - 1}{csc ^{2} ( x )}

Multiply:

1 \cdot csc^{2} (x) = csc^{2} (x)

= \frac{csc ^{2} ( x ) - 1}{csc ^{2} ( x )}

= \frac{\frac{c s c ^{2} ( x ) - 1}{c s c ^{2} ( x )} csc ^{2} ( x )}{1}

Apply the fraction rule:

\frac{a}{1} = a

= \frac{csc ^{2} ( x ) - 1}{csc ^{2} ( x )} csc^{2} (x)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( csc ^{2} ( x ) - 1 ) csc ^{2} ( x )}{csc ^{2} ( x )}

Cancel the common factor:

csc^{2} (x)

= csc^{2} (x) - 1

csc^{2} (x) - 1

csc^{2} (x) - 1

We showed that the two sides could take the same form

\Rightarrow True

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Is csc^2(x)-1=(cos(x))/(tan(x)sin(x)) ?
The answer to whether csc^2(x)-1=(cos(x))/(tan(x)sin(x)) is True